# Transformer Calculation Units Check

#### MrAl

Joined Jun 17, 2014
11,487
Hello there,

I figured @WBahn would appreciate this as he always talks about how important the specifying of units is.

This was found on another website and this 'manual' or whatever it was had a gross error.
Using that formula for N would result in a core requirement over 6 times as really needed. Imagine using a transformer with cross sectional area of 6 square inches instead of the required 1 square inch. Crazy.
The units of cm^2 agrees with Thomas and Skinner as well as other references.

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#### WBahn

Joined Mar 31, 2012
30,065
Hello there,

I figured @WBahn would appreciate this as he always talks about how important the specifying of units is.

This was found on another website and this 'manual' or whatever it was had a gross error.
Using that formula for N would result in a core requirement over 6 times as really needed. Imagine using a transformer with cross sectional area of 6 square inches instead of the required 1 square inch. Crazy.
The units of cm^2 agrees with Thomas and Skinner as well as other references.
I can't read what the exponent is in the numerator, but this whole notion of having to specify what the units are for the terms in the equation shows a lack of proper treatment for units -- the variables carry their own units.

The basic equation is simply

$$E \; = \; N \omega A B$$

N is the number of turns
w is the frequency
A is the cross sectional area
B is the magnetic flux density

Done.

If we want to assume that everything is nice an sinusoidal, the we can make this a bit more convenient:

$$E_{pk} \; = \; N \omega A B_{pk} \\ E_{pk} \; = \; \sqrt{2}E_{rms} \\ E_{rms} \; = \; \frac{N \omega A B_{pk}}{\sqrt{2}}$$

If we want to work with cyclical frequency instead of radian frequency, then we can go further:

$$\omega \; = \; \left( 2\pi \; \frac{rad}{cyc} \right) f \\ E_{rms} \; = \; \frac{2\pi \; rad}{\sqrt{2} \; cyc} N f A B_{pk} \\ E_{rms} \; = \; 4.44 \cdot N f A B_{pk} \; \frac{rad}{cyc}$$

Admittedly, the conversion between cyclical frequency and radian frequency seldom properly track the units and is usually just written as w = 2πf (and this is one that I admit to being sloppy about and just dealing with mentally most of the time), but here I am being rigorous.

So now solve for N and plug in values for the rest using whatever appropriate units you want. Use cycles per fortnight, area in hectares, or B in ohm-electrons. You will end up with a valid value for the RMS potential across the coil, it will merely be expressed in units that we have no intuitive grasp for, just as most people would find grasping with the area of a room being expressed in ft·m difficult. But it's valid. If you want to express it in a convenient unit of measure, just do the units conversions.

#### Ian0

Joined Aug 7, 2020
9,826
Hurrah for the SI system!
Some of our politicians want to revert to imperial units. Will my electricity bill be then in British Thermal Units or Foot Pounds?

But seriously, it's important to know where the mysterious 4.44 comes from. Otherwise you still use it when dealing with high-frequency transformers with square-wave drive, it will give you the wrong answer.

#### MrAl

Joined Jun 17, 2014
11,487
I can't read what the exponent is in the numerator, but this whole notion of having to specify what the units are for the terms in the equation shows a lack of proper treatment for units -- the variables carry their own units.

The basic equation is simply

$$E \; = \; N \omega A B$$

N is the number of turns
w is the frequency
A is the cross sectional area
B is the magnetic flux density

Done.

If we want to assume that everything is nice an sinusoidal, the we can make this a bit more convenient:

$$E_{pk} \; = \; N \omega A B_{pk} \\ E_{pk} \; = \; \sqrt{2}E_{rms} \\ E_{rms} \; = \; \frac{N \omega A B_{pk}}{\sqrt{2}}$$

If we want to work with cyclical frequency instead of radian frequency, then we can go further:

$$\omega \; = \; \left( 2\pi \; \frac{rad}{cyc} \right) f \\ E_{rms} \; = \; \frac{2\pi \; rad}{\sqrt{2} \; cyc} N f A B_{pk} \\ E_{rms} \; = \; 4.44 \cdot N f A B_{pk} \; \frac{rad}{cyc}$$

Admittedly, the conversion between cyclical frequency and radian frequency seldom properly track the units and is usually just written as w = 2πf (and this is one that I admit to being sloppy about and just dealing with mentally most of the time), but here I am being rigorous.

So now solve for N and plug in values for the rest using whatever appropriate units you want. Use cycles per fortnight, area in hectares, or B in ohm-electrons. You will end up with a valid value for the RMS potential across the coil, it will merely be expressed in units that we have no intuitive grasp for, just as most people would find grasping with the area of a room being expressed in ft·m difficult. But it's valid. If you want to express it in a convenient unit of measure, just do the units conversions.
Hi,

I am not sure what you are saying with that first sentence except for the exponent which I have fixed in the new drawing. The exponent is an '8' which they got right, and any other single digit exponent will not correct the original problem.

In many of these papers they do not give the entire derivation just the formula. In that case the units have to be specified.
This formula is very common and shown in many manuals on transformers such as by Thomas and Skinner and Magnetics Inc. They always give the units for the area correctly though as square centimeters unless they change a constant that allows the use of square inches. The two constants for using square centimeters are always 1e8 and 4.44 (sine waves of course). In the paper, they mistakenly gave the units for the area as square inches but to do that one of those constants has to be changed, and the usual method is to lump them into one constant that includes those two and the conversion constant for inches to centimeters which is 6.45 and they did not do that.

As per the other reply, yes for square waves that 4.44 is changed to a lower value like 4.

Obviously there is a lot more to designing a transformer like this such as window area, surface area, thermal conductivity inside to outside, etc., and of course a good test for a new design. The usual test in places I've worked is an overnight run to see if the transformer melts down. If it gets too hot it melts the insulation between windings and often causes a short which blows a fuse. The insulation ends up making a big blob on the bench before the fuse blows. That of course indicates the design needs a revision.

Interesting, they did not specify the units for Bmax either which would be in Gausses.

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#### WBahn

Joined Mar 31, 2012
30,065
In many of these papers they do not give the entire derivation just the formula. In that case the units have to be specified.
Not if they treat units properly. That's the point.

Look at formulas we see and use all the time, such as

C = 2πR
A = πR²
A = 4πR²
V = 4πR^3/3
d = v*t
x = Xo + Vo·t + a·t²

Why don't we need to specify units in any of those?

Because the parameters carry their units.

Now, what if, for convenience, we want to have a formula in which we measure the radius of a sphere in inches but we need the volume to be in cc (cubic centimeters)?

Fine, then multiply the equation by the appropriate constant -- but that constant has units! You can't just throw them away and still claim that you are treating units properly.

$$V \; = \; \left( \frac{4}{3}\pi R^3 \right) \left( \frac{2.54 \; cm}{1 \; in} \right)^3 \\ V \; = \; 68.6 \; \frac{cc}{in^3} R^3$$

There is no need to specify that R now has to be in inches or that the volume will be in cubic centimeters. Not only does the equation make it obvious that this is the preferred measurements for that use, but it also in no way limits the equation to those units. Give someone that equation who is designing a big concrete sphere for a sculpture and knows that the radius of the sphere is 2.75 meters but needs to order the concrete in cubic yards, and there is no need for them to pour through books or the Internet looking for some formula with a magic number out front and a note saying that R is in meters and V is in cubic yards. Just take the perfectly fine equation above and track the units properly.

$$V \; = \; 68.6 \; \frac{cc}{in^3} R^3 \\ V \; = \; 68.6 \; \frac{cc}{in^3} \left( 2.75 \;m \right)^3 \\ V \; = \; 1,427 \; \frac{cc \cdot m^3}{in^3} \\$$

That's a perfectly valid volume, given in units of volume (namely length-cubed). That it isn't what we want is not a problem -- we do the exact same thing that we do if we want to convert something from acres to square miles, namely apply unit conversions to get from what we have to what we want.

$$V \; = \; 1,427 \; \frac{cc \cdot m^3}{in^3} \; \left( \frac{1 \; cm^3}{1 \; cc} \right) \; \left( \frac{100 \; cm}{1 \; m} \right)^3 \; \left( \frac{1 \; in}{2.54 \; cm} \right)^6 \; \left( \frac{1 \; yd}{36 \; in} \right)^3 \\ V \; = \; 113.9 \; yd^3$$

Which is what you get if you first convert 2.75 m to yards and plug that into the basic equation for the volume of a sphere.

This formula is very common and shown in many manuals on transformers such as by Thomas and Skinner and Magnetics Inc. They always give the units for the area correctly though as square centimeters unless they change a constant that allows the use of square inches. The two constants for using square centimeters are always 1e8 and 4.44 (sine waves of course). In the paper, they mistakenly gave the units for the area as square inches but to do that one of those constants has to be changed, and the usual method is to lump them into one constant that includes those two and the conversion constant for inches to centimeters which is 6.45 and they did not do that.
And they did not do that because they are sloppy with their units. They did EXACTLY what I always preach about -- they threw away the units and then tacked on the units (via a note) that they expected (although hoped, prayed, wished, and guessed are more appropriate verbs) the result to have at the end, instead of tracking the units and letting them take care of themselves and revealing what the units actually are at the end. And, as a direct result, they threw away the golden opportunity to catch their mistake right at the point where they made it and fix it right then and there before cementing their error for all the world to see in a publication, and doing it in such a way that the people that rely on their sloppy work can't easily catch it and, instead, will blindly use it and wonder why their designs are piss poor failures, possibly even causing damage, injuries, or deaths.

#### MisterBill2

Joined Jan 23, 2018
18,538
In several college classes I had instructors who were very demanding about including units. And the explanations were similar. Certainly it is a very simple check that the expression is VALID, although it might be incorrect.
We see a lack of units in the "over unity efficiency" posts quite a bit. A simple way to go astray.

#### MrAl

Joined Jun 17, 2014
11,487
Not if they treat units properly. That's the point.

Look at formulas we see and use all the time, such as

C = 2πR
A = πR²
A = 4πR²
V = 4πR^3/3
d = v*t
x = Xo + Vo·t + a·t²

Why don't we need to specify units in any of those?

Because the parameters carry their units.

Now, what if, for convenience, we want to have a formula in which we measure the radius of a sphere in inches but we need the volume to be in cc (cubic centimeters)?

Fine, then multiply the equation by the appropriate constant -- but that constant has units! You can't just throw them away and still claim that you are treating units properly.

$$V \; = \; \left( \frac{4}{3}\pi R^3 \right) \left( \frac{2.54 \; cm}{1 \; in} \right)^3 \\ V \; = \; 68.6 \; \frac{cc}{in^3} R^3$$

There is no need to specify that R now has to be in inches or that the volume will be in cubic centimeters. Not only does the equation make it obvious that this is the preferred measurements for that use, but it also in no way limits the equation to those units. Give someone that equation who is designing a big concrete sphere for a sculpture and knows that the radius of the sphere is 2.75 meters but needs to order the concrete in cubic yards, and there is no need for them to pour through books or the Internet looking for some formula with a magic number out front and a note saying that R is in meters and V is in cubic yards. Just take the perfectly fine equation above and track the units properly.

$$V \; = \; 68.6 \; \frac{cc}{in^3} R^3 \\ V \; = \; 68.6 \; \frac{cc}{in^3} \left( 2.75 \;m \right)^3 \\ V \; = \; 1,427 \; \frac{cc \cdot m^3}{in^3} \\$$

That's a perfectly valid volume, given in units of volume (namely length-cubed). That it isn't what we want is not a problem -- we do the exact same thing that we do if we want to convert something from acres to square miles, namely apply unit conversions to get from what we have to what we want.

$$V \; = \; 1,427 \; \frac{cc \cdot m^3}{in^3} \; \left( \frac{1 \; cm^3}{1 \; cc} \right) \; \left( \frac{100 \; cm}{1 \; m} \right)^3 \; \left( \frac{1 \; in}{2.54 \; cm} \right)^6 \; \left( \frac{1 \; yd}{36 \; in} \right)^3 \\ V \; = \; 113.9 \; yd^3$$

Which is what you get if you first convert 2.75 m to yards and plug that into the basic equation for the volume of a sphere.

And they did not do that because they are sloppy with their units. They did EXACTLY what I always preach about -- they threw away the units and then tacked on the units (via a note) that they expected (although hoped, prayed, wished, and guessed are more appropriate verbs) the result to have at the end, instead of tracking the units and letting them take care of themselves and revealing what the units actually are at the end. And, as a direct result, they threw away the golden opportunity to catch their mistake right at the point where they made it and fix it right then and there before cementing their error for all the world to see in a publication, and doing it in such a way that the people that rely on their sloppy work can't easily catch it and, instead, will blindly use it and wonder why their designs are piss poor failures, possibly even causing damage, injuries, or deaths.
Well then you are going to hate this next image taken from one of the handbooks
More units quoted than you can ever hope for.
They are trying to make it as simple as possible for the designers.

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#### WBahn

Joined Mar 31, 2012
30,065
Well then you are going to hate this next image taken from one of the handbooks
No surprise there. Of course I hate it whenever I see sloppy units. Just because sloppiness with units is rampant and everywhere does NOT make it good, or correct. It merely makes it that much more pathetic. What I really hate is when people that are sloppy with units claim that they are being careful with units. But the worst are the textbook authors who carry on about the importance of units in Chapter 1, and then proceed to be sloppy with units throughout the rest of the work and just take the typical approach of ignoring the units and tacking on something out of thin air at the end.

I've pointed out numerous times why I believe this is so rampant. It starts with the math teachers in grade school when kids are introduced to "story problem" -- they are taught to pull the numbers out of a problem, do some arithmetic on them, and then write down an answer, possibly with the expectation that the answer have units tacked onto it. So right from the get-go, kids are primed that this is how you deal with units. By the time they get to high school and, perhaps, take a decent physics or chemistry class, this is so engrained that they give lip service to units in those classes just to keep from losing points, but seldom let it influence how they treat units when they aren't forced to. At the college level it is even worse in most engineering disciplines. They are being taught by professors that have never worked in the real world, using textbooks written by authors that have never worked in the real world, and so all of their bad habits are firmly reinforced.

#### MrAl

Joined Jun 17, 2014
11,487
Hi,

Yeah, and sometimes they just state the units beforehand.

For my personal preferences, I find it faster to read if the units are stated clearly. That way I do not have to backtrack to find out what units they are using, or start to apply physics all over again just to figure out the units. In plain design formulas, I always try to remember to state the units explicitly. If I think the reader is a relative newcomer, I'll even state the brief definition of 'pi' as 3.14159... and sometimes 'e' and often 'w' as 2*pi*f with f in Hertz.

#### MisterBill2

Joined Jan 23, 2018
18,538
More than once in my career, when an answer did not look reasonable, going back thru the math and verifying all of the units was the fast way to find the error, usually losing one term someplace.

#### MrAl

Joined Jun 17, 2014
11,487
More than once in my career, when an answer did not look reasonable, going back thru the math and verifying all of the units was the fast way to find the error, usually losing one term someplace.
Yes and here are errors all over the web. You have to start from scratch, and some of that stuff can be really complicated.
I ended up starting from scratch many times. It's a big pain. I think because there is probably very little error checking like we used to see with physical books. I found errors in technical books too, but not as bad as on the web.